Question

About 38% of people who get their eyes examined are found to be farsighted (unable to see objects close-up). What is the probability of an ophthalmologist examining six people until he finds the first patient who is farsighted? (0.62)5(0.38)1 (0.38)6 (0.38)5(0.62)1 (0.62)6 1 − (0.38)5

Answer 1

Answer:

1) Probability of not getting a patient with farsightedness = 1-0.38 = 0..62.

Probability of examining 6 patients until a patient with farsightedness is examined = The Probability that the first six patients do not have farsightedness = 0.62*0.62*0.62*0.62*0.62*0.62 = 0.626. So option 4 is the correct answer.

1 - (0.38)5

Answer 2

The probability of an ophthalmologist examining six people until he finds the first patient who is farsighted is (0.62)⁵(0.38)¹.

What is Probability?

Probability helps us to know the chances of an event occurring.

$\rm Probability=\dfrac{Desired\ Outcomes}{Total\ Number\ of\ outcomes\ possible}$

Given that 38% of people who get their eyes examined are found to be farsighted. Therefore, the probabilities of getting a farsighted patient and not getting a farsighted patient can be written as,

The probability of a patient being farsighted, p = 0.38

The probability of a patient not being farsighted, q = 1 - 0.38 = 0.62

Now, the probability of an ophthalmologist examining six people until he finds the first patient who is farsighted can be written as,

Probability = q × q × q × q × q × p

= q⁵ × p

= (0.62)⁵(0.38)¹

Hence, the probability of an ophthalmologist examining six people until he finds the first patient who is farsighted is (0.62)⁵(0.38)¹.

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